# Coefficient of x^3

1. May 15, 2008

### Stacyg

Find the coefficient of x^3 in the expansion of (2x^2-3/x)^3

I know how to do simple coefficients using pascalles triangle but I really don't know how to do this.
Any help would be much appreciated.

2. May 15, 2008

### Tedjn

Write it as

$$(2x^2 - 3x^{-1})^3$$

From Pascal's triangle, you know how to expand

$$(a+b)^n$$

What can you replace with a and what can you replace with b?

3. May 15, 2008

### rock.freak667

$$(2x^2-\frac{3}{x})^3$$

$$(\frac{1}{x}(2x^3-3))^3$$

4. May 15, 2008

### Tedjn

Oh, that's a nice way of doing it :)

5. May 15, 2008

### rock.freak667

Usually (well for me), a binomial expansion is usually done with a variable and a constant.

as for $(a+b)^n$ is valid for $|\frac{b}{a}|<1$ But if a and b are variables, you'll have to do some fancy algebra to get the range for which it is valid.

6. May 15, 2008

### Dick

Why is it only valid in some range??? I also don't see why you need to factor the original. (a+b)^3=a^3+3*a^2*b+3*a*b^2+b^3. Just put a=2x^2 and b=(-3/x), figure out which term is the x^3 term and evaluate it.

7. May 15, 2008

### rock.freak667

That's what I was taught.."validity of a binomial"

8. May 15, 2008

### Dick

Got a reference? If you are thinking of the convergence of the infinite series for negative exponents, that is something to think about. But this is a positive exponent, the series is finite. There are no convergence issues.

9. May 15, 2008

### BrendanH

Besides, we're dealing with polynomials in the case of (a+b)^n

10. May 15, 2008

### Dick

Yeah, that's what I mean by finite series. You could also just forget about pascal's triangle and multiply it out. The power is only 3.